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Chapter 13: Problem 73

If an electron in \(\mathrm{H}\) -atom jumps from one orbit to other, its angular momentum doubles. The distance of electron from nucleus becomes ___times the initial distance. (a) 2 (b) 4 (c) \(\frac{1}{2}\) (d) \(\frac{1}{4}\)

Short Answer

Expert verified
The distance of the electron from the nucleus becomes 4 times the initial distance.

Step by step solution

01

Understand the Angular Momentum of an Electron in an Atom

According to the Bohr model of the hydrogen atom, the angular momentum of an electron in a stable orbit is quantized and is given by the relation \( L = n\hbar \), where \( n \) is the principal quantum number and \( \hbar \) is the reduced Planck's constant.
02

Relate Angular Momentum to the Principal Quantum Number

The problem states that the angular momentum doubles. If the initial angular momentum is \( L_1 = n_1\hbar \), then the final angular momentum \( L_2 \) will be twice that, or \( L_2 = 2n_1\hbar = n_2\hbar \). Therefore, \( n_2 = 2n_1 \). The principal quantum number has doubled.
03

Determine the Relationship Between Radius and Principal Quantum Number

In the Bohr model, the radius of the electron's orbit is directly proportional to the square of the principal quantum number, given by the relation \( r \propto n^2 \). Thus, if the principal quantum number doubles, the square of the principal quantum number quadruples.
04

Calculate the New Distance as a Multiple of the Initial Distance

Since the initial distance \( r_1 \propto n_1^2 \) and the final distance \( r_2 \propto n_2^2 \), and \( n_2 = 2n_1 \), we have \( r_2 \propto (2n_1)^2 = 4n_1^2 \). Therefore, the distance of the electron from the nucleus becomes 4 times the initial distance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Momentum Quantization
The exercise improvement suggests a need for clarification on the significance of quantization. To put it simply, angular momentum quantization restricts the electron's 'path' around the nucleus to certain allowed orbits. If an electron in a hydrogen atom jumps from one orbit to another, causing its angular momentum to double, we understand this as a shift from one quantized state to another with a higher principal quantum number.
Principal Quantum Number
The significance of the principal quantum number extends beyond the atom's current energy state. Quantum numbers contribute to the 'address' of an electron, much like the coordinates on a map, allowing us to predict and understand its behavior. The value of n leads to predictions about not only the electron's energy but also the distance from the nucleus and the likelihood of its location in a particular area around the nucleus. Such insights are fundamental to comprehending chemical bonding and the properties of elements.
Electron Orbital Radius
Understanding the radius of an electron's orbit is essential for visualizing the structure of the atom and the spatial distribution of electronic clouds. These orbital radius calculations help in visualizing the atom as a three-dimensional structure and are key to explaining the atom's chemical reactivity and interaction with light. With the solution to our exercise indicating that the electron's distance from the nucleus becomes four times the initial distance, it illustrates a fundamental principle of the electron's quantized movement in atomic structure.

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Most popular questions from this chapter

If the numbers of orbitals of a particular type were \((3 l+1)\), but spin quantum numbers were only \(+1 / 2\) and \(-1 / 2\), then \(\mathrm{d}\) -type orbitals will contain a maximum of electrons. (a) 10 (b) 14 (c) 7 (d) 5

A ruby laser produces radiations of wavelength \(662.6 \mathrm{~nm}\) in pulses whose duration are \(1.0 \times 10^{-9} \mathrm{~s}\). If the laser produces \(0.36 \mathrm{~J}\) of energy per pulse, how many photons are produced in each pulse? (a) \(1.2 \times 10^{9}\) (b) \(1.2 \times 10^{27}\) (c) \(1.2 \times 10^{18}\) (d) \(1.2 \times 10^{15}\)

A certain laser transition emits \(6.0 \times 10^{15}\) quanta per sec per square metre of \(\lambda=662.6 \mathrm{~nm} .\) What is the power output in joule per sec per square metre? (a) \(1.8 \times 10^{-3}\) (b) \(6.626 \times 10^{-4}\) (c) \(1.8 \times 10^{3}\) (d) \(6.626 \times 10^{-12}\)

Number of electrons having \(m=0\) for sodium atom is (a) 2 (b) 5 (c) 7 (d) 3

Two carbon discs, \(1.0 \mathrm{~g}\) each, are \(1.0 \mathrm{~cm}\) apart have equal and opposite charges. If force of attraction between them is \(10^{-5} \mathrm{~N}\), the ratio of excess electrons to the total atoms on the negatively charged disc is \(\left(N_{\mathrm{A}}=6 \times 10^{23}\right)\) (a) \(2.4 \times 10^{-12}: 1\) (b) \(10^{-12}: 2.4\) (c) \(10^{12}: 2.4\) (d) \(2.4: 10^{12}\)
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